Екі айнымалысы бар сызықтық емес теңдеулер жүйесін шешу. Алгебра, 9 сынып, қосымша материал. 1 сабақ.

9.3

I would read about the history of the subject I was taking . . . so I had a more comprehensive view of the subject than my fellow students. Many times I would walk up to the exam with some classmate and say, “I will review you for the exam, I’ll ask you some questions,” and he would give me the answers he had studied. I’d go in, take the exam, and get 20 percent more than he did. He’d be so full of the subject, he couldn’t see the woods for the trees.

I. I. Rabi

[–5, 5] by [–3, 3]

9.3Systems of Nonlinear Equations509

SYSTEMS OF NONLINEAR EQUATIONS

Gaussian elimination and matrix methods are well-suited for systems of linear equations. However, we often must deal with systems that include nonlinear equations. Such systems can sometimes be difficult to solve, even with the aid of technology. Several of the ideas we have used to solve systems of linear equations have applicability to nonlinear systems. In particular, we often use the method of substitution, which is a special case of eliminating a variable. If we can graph the equations, we can make use of technology as well.

H

x2 1 y2 5 5

cEXAMPLE 1 A nonlinear system For the system₂ , (a) dex 5 y 2 3

scribe three different strategies for solving, and (b) show that two of your strategies in part (a) yield the same solution set.

Solution

• (i) Since the second equation is already solved for x, we can substitute y² 2 3 for x in the first equation and solve the resulting equation for y.

x2 1 y2 5 52

• • Writing the system in the form₂, we can eliminate y by

x 2 y 523

adding equations ~replace E₂ by E₂ 1 E₁!.

• • We can solve both equations for y and graph four equations: y 5

6Ï5 2 x², y 56Ïx 1 3. Then use graphical methods.

• (i) Substituting for x, the first equation becomes

~y² 2 3!² 1 y² 5 5,ory⁴ 2 5y² 1 4 5 0.

Factoring, ~y² 2 1!~y² 2 4! 5 0, so y 561 or y 562. The corresponding x-values are 22 and 1. The solution set consists of the four points $~22, 1!, ~22, 21!, ~1, 2!, ~1, 22!%.

• • Adding equations gives x² 1 x 5 2, with solutions x 522, 1. Substituting each x-value into either of the original equations and solving for y gives the same four points as the solution set.

• • Graphing the four equations y 56Ï5 2 x², y 56Ïx 1 3 on the same screen gives something like Figure 5. In a decimal window we can read the coordinates of the same four points exactly, but in general we would have to settle for approximations. b

cEXAMPLE2 Solving with graphs Solve the system of equations and show the solutions graphically. x 1 2y 524

FIGURE 5y 5 x² 2 2x 2 3

510Chapter 9Systems of Equations and Inequalities

Strategy: Eliminate either x or y by substituting from the first equation into the second, and then solving a quadratic.

y = x² – 2x – 3 ^y

FIGURE 6

y

Solution

Follow the strategy. Solving the first equation for x gives x 522y 2 4. Substitute into the second equation and solve for y.

y 5 ~22y 2 4!² 2 2~22y 2 4! 2 3 y 5 4y² 1 16y 1 16 1 4y 1 8 2 3 4y² 1 19y 1 21 5 0

~4y 1 7!~y 1 3! 5 0

Therefore, y 52 or y 523. Now use x 522y 2 4 to get the corresponding values of x. For y 52, x 52; for y 523, x 5 2.

 The graph of the first equation is a line and that of the second is a parabola that opens upward with vertex at ~1, 24!. See Figure 6. The points of intersection are ~2, 2! and ~2, 23!, which correspond to the two solutions. b

cEXAMPLE 3 Complex solutionSolve the system of equations and interpret the solution graphically.

2x 1 y 5 3 x² 1 y 5 1

Solution

Solve the second equation for y to get y 5 1 2 x². Substitute into the first equation and solve for x.

2x 1 ~1 2 x²! 5 3

x² 2 2x 1 2 5 0

To solve the quadratic equation, apply the quadratic formula.

 2 6Ï4 2 8 2 6Ï24 2 6 2i x 55 5 5 1 6 i.

222

We find the coordinates of the points of intersection, if any, of the parabola and the line by solving the system for real number solutions only. Since we have imaginary number solutions, there are no points of intersection, as we see from the graphs in

FIGURE 7Figure 7. b

cEXAMPLE 4 Eliminate variableFind the solution set for the system of equations

x2 2 y2 5 3

2x² 1 y² 5 9.

Solution

Eliminate y by adding the two equations and then solve for x.

3x² 5 12x² 5 4x 562.

To get the corresponding values of y, substitute 2 or 22 into either of the given equations, say the first.

4 2 y² 5 3y² 5 1y 561.

This gives four solutions. The solution set is $~2, 1!, ~2, 21!, ~22, 1!, ~22, 21!%. b

9.3Systems of Nonlinear Equations511

ln x² 5 ln@2~4 2 x!# x² 5 2~4 2 x! x² 1 2x 2 8 5 0

~x 2 2!~x 1 4! 5 0

x Therefore, we get two possible solutions for x: x 5 2 or x 524, but 24 is not a solution. See the Strategy. To find the corresponding value of y use either of the given equations, say y 5 2 ln x. For x 5 2, y 5 2 ln 2 5 ln 4. The graphs of the two equations intersect at only one point ~2, ln 4! < ~2, 1.39!. See Figure 8. b

FIGURE 8

EXERCISES 9.3

Check Your Understanding

Exercise 1–6True or False. Give reasons.

• The system of equations

Ï2x 2Ï5y 5 4

Ï3x 1Ï7y 5 3

is a nonlinear system.

• The graphs of y 5 x² and y 52x 2 1 intersect at two points.
• The graphs of y 5 _ x _ and y 5 x have in common only one point, ~0, 0!.
• The system of equations

 x 2 y 5 0 x2 1 y2 5 8

has exactly two solutions.

• The system

 x 2 2y 2 1 5 0 x² 1 y² 1 1 5 0

has no real solutions.

• The system

x² 1 y 5 0

x² 2 y 5 0

has no real solutions.

Exercises 7–10 Fillintheblank sothattheresultingstatement is true.

• The graphs of y 5 x² 2 1 and y 5 1 2 x² intersect at

.

• The graphs of x² 1 y² 5 25 and 4x 1 3y 5 0 intersect at.
• The graphs of _ x _ 1 y 5 0 and y 1 2 5 0 intersect at

.

• The graphs of y 5 _ x _ 2 1 and y 5 1 2 _ x _ intersect at.

Develop Mastery

Exercises 1–16 Solve, Draw Graphs Find all pairs of real numbers x, y that satisfy the system of equations. Draw graphs and show points of intersection (if any).

1. y 5 3x 1 42. 2x 2 y 1 2 5 0 y 5 x²x² 1 y² 5 169

3.3x 1 y 5 04. 2x 2 y 522

2x² 1 4x 1 y 5 0xy 5 4

5. 2x 1 3y 5236.5x 2 y 5 10 xy 523x² 1 x 2 y 5 6

7. 2x 2 y 5 08.x 1 y 5 2 x² 2 y 523x² 1 y² 5 2

9. 3x 2 y 5 510. y 5 x² 2 4x 1 4

x² 1 y² 5 25y 522x² 1 x 1 16

512Chapter 9Systems of Equations and Inequalities

Exercises 17–30 Nonlinear Systems Solve the system of equations. If results involve irrational numbers, give approximations rounded off to two decimal places.

• x² 1 y² 2 xy 5 3 x2 1 y2 5 5
• 2x² 1 5xy 1 3y² 5 4 xy 522

Exercises 31–34 Trigonometric Functions Solve the system of equations. Assume that 0 # x # 2p; for Exercises 33 and 34, 0 # y # 2p.

• sin x 2 y 5 032.sin x 1 y 5 0 cos x 2 y 5 0sin 2x 2 y 5 0
• 2 sin x 1 cos y 5 2 sin x 2 cos y 520.5
• sin x 1 cos y 5 0

2 sin x 2 4 cos y 5 3Ï2

Exercises 35–36 Absolute Values Solve the system of equations. (Hint: How could you graph an equation involving _ y _?!

• 3_ x _ 2 2_ y _ 522 _ x _ 1 3_ y _ 5 14
• 2_ x _ 2 3_ y _ 5 0

4_ x _ 1 3_ y _ 5 18

Exercises 37–42Nonlinear Systems

37. 6e^x 2 e^y 5 138. e^x 1 2e^y 5 8

3ex 1 ey 5 82ex 2 ey 5 1

39. ln x 1 ln y 5 0 40. ln x 1 ln y 5 0 2 ln x 1 ln y 5 1 3 ln x 1 4 ln y 5 2

• x 1 y 1 _ x _ 5 9 x 2 y 1 _ y _ 5 12

(Hint: What are the possible values of x 1 _ x _?)

• x 1 y 1Ïx² 5 6 x 1Ïy² 2 y 5 8

Exercises 43–46Rectangles

• The perimeter of a rectangle is 40 cm and the area is 96 cm². Find the dimensions of the rectangle.
• Find the dimensions of a rectangle that has a diagonal of length 13 cm and a perimeter of 34 cm.
• One side of a rectangle is 3 cm longer than twice the shorter side, and the area is 230 cm². Find the perimeter of the rectangle.

• A rectangle is inscribed in a circle of radius Ï10. If the area of the rectangle is 16, find its dimensions.

Exercises 47–48 Line Through Intersections

• Find an equation for the line that passes through the points of intersection of the graphs of y 5 x² 1 2x and y 52x².
• Find an equation for the line that passes through the points of intersection of the graphs of y 5 x² 2 4x 2 5 and y 52x² 1 2x 1 3.
• An altitude of a triangle is twice as long as the corresponding base and the area of the triangle is 36 cm². Find the altitude and the base. Does the given information determine a unique triangle? Suppose the problem states that one of the other sides is 4Ï10. What is the perimeter of the triangle?
• Explore Find all pairs of real numbers (if any) such that

(a) their difference is 1 and their product is 1. (b) their sum is 1 and their product is 1.

(c) their difference is 1 and their quotient is 1. (d) their sum is 1 and their quotient is 1.